# How to understand the continuity equation $i\omega \rho=\nabla\cdot\mathbf{J}$?

Suppose that $\mathbf{J}$ is alternating current and its intrinsic frequency is $\omega$, i.e. $\mathbf{J}(\mathbf{r}',t)=\mathbf{J}(\mathbf{r}')e^{-i\omega t}$ , and $\rho$ is charge density. Then my textbook(one Chinese book, no English version, it's called 《电动力学》and written by 郭硕鸿) says they satisfy the law of conservation of charge, so that is $$i\omega\rho=\nabla\cdot\mathbf{J} \tag{1}$$

I know the continuity equation is $$-\frac{\partial \rho}{\partial t}=\nabla\cdot\mathbf{J}$$ and it could be derived from the fourth Maxwell's equation. I have tried my best to deduce the equation $(1)$, but I faild.

Here, $\nabla\cdot\mathbf{J}(r',t)=e^{-i\omega t}\nabla\cdot\mathbf{J}(r')$, and from that equation I know $-e^{i\omega t}\frac{\partial \rho}{\partial t}=\nabla\cdot\mathbf{J}(r',t)$, however, I don't know how to do from this.

• Try $\rho(r',t) = \rho(r') e^{-i \omega t}$. – Muphrid Jan 11 '16 at 1:27
• @Muphrid I tried and succeeded, Thank you. And you could write your comment in the answer area, then I will accept your answer. – Wang Yun Jan 11 '16 at 1:31

Like with $J$, it's assumed that $\rho(r',t) = \rho(r') e^{-i\omega t}$--that both the current and charge density oscillate at the same frequency.